Question

How many grams of a liquid of specific heat $0.2$ at a temperature $40^{\circ} C$ must be mixed with $100\, gm$ of a liquid of specific heat of $0.5$ at a temperature $20^{\circ} C$, so that the final temperature of the mixture becomes $32^{\circ} C$

• A. 175 gm
• B. 300 g
• C. 295 gm
• D. 375 g

Answer 1

Answer: (D)

Temperature of mixture
$\theta=\frac{m_{1} c_{1} \theta_{1}+m_{2} c_{2} \theta_{2}}{m_{1} c_{1}+m_{2} \theta_{2}}$
$\Rightarrow 32=\frac{m_{1} \times 0.2 \times 40+100 \times 0.5 \times 20}{m_{1} \times 0.2+100 \times 0.5}$
$\Rightarrow m_{1}=375\, gm$